XAT 2022 — QA & DI Question 6

Considering the two numbers to a, b :
We were given that:

a³ + b³ = 128

$\frac{1}{a^3}$ + $\frac{1}{b^3}$ = 2

$\left(\frac{a^3+b^3}{a^3.b^3}\right)$ = 2 = $\frac{128}{k}$

k = 64.

Hence a³ . b³ = 64

a*b = 4 and a² .b² = 16